If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?

If the solenoid in the above question is free to turn about the vertic

1.	If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?
Answer» Magnetic field strength, `B=0.25T` M agnetic moment, `M=0.6T^(-1)` The angle `theta`, between the axis of the solenoid and the direction of the applied field is `30^(@)`. Therefore, the torque acting on the solenoid is given as: `tau=MB sin theta` `=0.6xx0.25sin30^(@)` `=7.5xx10^(-2)J`

Discussion

No Comment Found

Related InterviewSolutions

The variation of magnetic susceptibility `(chi)` with absolute temperature `T` for a ferromagnetic material is
Point out the best representation of relation between magnetic susceptibility `(Chi)` and temperature (T) for a paramagnetic materialA. B. C. D.
The hysteresis loss for a specimen of iron weighing 15kg is equivalent at `300Jm^(-3)cycle^(-1)` Find the loss of energy per hour at 25 cycle `s^(-1)`. Density of iron is `7500kg m^(-3)`
Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of `60^(@)` and `45^(@)` respectively. The ratio of the number of turns in the coils is
A bar magnet of pole strength 10A-m is cut into two equal parts breathwise. The pole strength of each magnet isA. 5A-mB. 10A-mC. 15AD. 15A-m
A bar magnet of magnetic moment M and pole strength m is broken in two equal part along the magnetic axis . The magnetic moment and pole strength of each part areA. `(M)/(2) , (m)/(2)`B. `(M)/(2)` , mC. `M , (m)/(2)`D. M , m
Assertion : When a bar magnet is freely suspended, it points in the north-south direction. Reason : The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to north.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is ture but reason is false.D. If both assertion and reason are false.
The time period of oscillation of a freely suspended bar magnet with usual notations is given byA. `T=2pisqrt(I/(MB_(H)))`B. `T=2pisqrt((MB_(H))/I)`C. `T=sqrt(I/(MB_(H)))`D. `T=2pisqrt(B_(H)/(MI))`
The couple acting on a magnet of length 10cm and pole strength 15A-m, kept in a field of `B=2xx10^(-5)`, at an anlge of `30^(@)` isA. `1.5xx10^(-5)N-m`B. `1.5xx10^(-3)N-m`C. `1.5xx10^(-2)N-m`D. `1.5xx10^(-6)N-m`
A magnet of magnetic moment `M` is situated with its axis along the direction of a magnetic field of strength `B`. The work done in rotating it by an angle of `180^(@)` will beA. `-MB`B. `+MB`C. `0`D. `+2MB`

If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment